2017乌鲁木齐网络赛 J题 Our Journey of Dalian Ends ( 最小费用最大流 )

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e4 + 10;
const int  INF = 0x3f3f3f3f;
struct st{ int from, to, w; }arr[10010];
map<string, int> mp;
int id;

struct Edge
{
    int from,to,cap,flow,cost;
    Edge(int u,int v,int ca,int f,int co):from(u),to(v),cap(ca),flow(f),cost(co){};
};

struct MCMF
{
    int n,m,s,t;
    vector<Edge> edges;
    vector<int> G[maxn];
    int inq[maxn];//是否在队列中
    int d[maxn];//距离
    int p[maxn];//上一条弧
    int a[maxn];//可改进量

    void init(int n)//初始化
    {
        this->n=n;
        for(int i=0;i<=n;i++)
            G[i].clear();
        edges.clear();
    }

    void AddEdge(int from,int to,int cap,int cost)//加边
    {
        edges.push_back(Edge(from,to,cap,0,cost));
        edges.push_back(Edge(to,from,0,0,-cost));
        int m=edges.size();
        G[from].push_back(m-2);
        G[to].push_back(m-1);
    }

    bool SPFA(int s,int t,int &flow,int &cost)//寻找最小费用的增广路，使用引用同时修改原flow,cost
    {
        for(int i=0;i<n;i++)
            d[i]=INF;
        memset(inq,0,sizeof(inq));
        d[s]=0;inq[s]=1;p[s]=0;a[s]=INF;
        queue<int> Q;
        Q.push(s);
        while(!Q.empty())
        {
            int u=Q.front();
            Q.pop();
            inq[u]--;
            for(int i=0;i<G[u].size();i++)
            {
                Edge& e=edges[G[u][i]];
                if(e.cap>e.flow && d[e.to]>d[u]+e.cost)//满足可增广且可变短
                {
                    d[e.to]=d[u]+e.cost;
                    p[e.to]=G[u][i];
                    a[e.to]=min(a[u],e.cap-e.flow);
                    if(!inq[e.to])
                    {
                        inq[e.to]++;
                        Q.push(e.to);
                    }
                }
            }
        }
        if(d[t]==INF) return false;//汇点不可达则退出
        flow+=a[t];
        cost+=d[t]*a[t];
        int u=t;
        while(u!=s)//更新正向边和反向边
        {
            edges[p[u]].flow+=a[t];
            edges[p[u]^1].flow-=a[t];
            u=edges[p[u]].from;
        }
        return true;
    }

    int MincotMaxflow(int s,int t)
    {
        int flow=0,cost=0;
        while(SPFA(s,t,flow,cost));
        return cost;
    }
}MM;

inline void init()
{
    mp.clear();
    mp["Shanghai"] = 1;///中转点上海
    mp["Dalian"] = 2;///起点大连
    mp["Xian"] = 3;///终点西安
    id = 4;
}

int main(void)
{
    int nCase;
    cin>>nCase;
    while(nCase--){
        init();
        int M;
        cin>>M;
        string From, To;
        int Weight;
        for(int i=1; i<=M; i++){
            cin>>From>>To>>Weight;
            if(!mp.count(From)) mp[From] = id++;
            if(!mp.count(To)) mp[To] = id++;
            arr[i].from = mp[From];
            arr[i].to = mp[To];
            arr[i].w = Weight;
        }

        int n = id-1;
        MM.init(2*n+1);
        MM.AddEdge(0, 2, 1, 0);
        MM.AddEdge(0, 3, 1, 0);
        for(int i=1; i<=n; i++)
            MM.AddEdge(i, i+n, 1, 0);
        for(int i=1; i<=M; i++){
            MM.AddEdge(arr[i].from+n, arr[i].to, 1, arr[i].w);
            MM.AddEdge(arr[i].to+n, arr[i].from, 1, arr[i].w);
        }

        printf("%d\n", MM.MincotMaxflow(0, 1));
    }
    return 0;
}